Question

解方程 （4X^2+3）（4X^2+2）=12 5/X+1 — 10/X+10 =2/3X-3 6（X^2+1/X^2) +5（X+1/X=38

Answer 1

1)
设4x^2 = t (t>0)
(t+3)(t+2) = 12
t^2 +5t -6=0
t-1)*(t+6)=0
t = 1, t = -6(舍去）
所以4x^2 = 1
x = 1/2 x = -1/2

2)
5/(x+1) -10/(x+10) = 2/(3x-3)

5(3x+3)(x+10)-10(x+1)(3x-3) = 2(x+1)(x+10)

17x^2 -143x - 160 = 0
(x+1)*(17x-160)=0
x = -1, x = 160/17
经检验，x = -1, x = 160/17都是原方程的根
所以原方程的根是x = -1, x = 160/17.

3)
看不清楚

追问

6（X^2+1/X^2) +5（X+1/X)=38

追答

设x+1/x =t t不为0
原方程可化为：
6(t^2 -2) +5t-38=0
6t^2 +5t -50 = 0
(2t-5)(3t+10)= 0
t=5/2 t = -10/3

所以
x+1/x = 5/2 x+1/x = -10/3
2x^2 - 5x +2 =0 或 3x^2 +10x +3 = 0
(2x-1)(x-2)=0 ( x+3）（3x+1) = 0
x = 1/2 x = 2 x = -3 x= -1/3

经检验： x = 1/2 x = 2 x = -3 x= -1/3都是原方程的根
所以原方程的根是：x = 1/2 x = 2 x = -3 x= -1/3。

Answer 2

1）（4X^2+3）（4X^2+2）=12
（4X^2+2+1）（4X^2+2）=12
把（4X^2+2）看成一个整体
(4x²+2)²+ (4x²+2)=12
4x²+2=3
x=±1/2
2) 无解
3)x=2或1/2 或 x=-3或-1/3